Question: What is the slope of the line tangent to $f(x) = 2x^{2}-3x-3$ at $x = 1$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(2(x+h)^{2}-3(x+h)-3) - (2x^{2}-3x-3)}{h}$ $ = \lim_{h \to 0} \frac{(2(x^{2}+2x h+h^{2})-3(x+h)-3) - (2x^{2}-3x-3)}{h}$ $ = \lim_{h \to 0} \frac{2x^{2}+4(x h)+2h^{2}-3x-3h-3-2x^{2}+3x+3}{h}$ $ = \lim_{h \to 0} \frac{4(x h)+2h^{2}-3h}{h}$ $ = \lim_{h \to 0} 4x+2h-3$ $ = 4x-3$ $ = (4)(1)-3$ $ = 1$